Search for a Range
Mar 3 '12
5829 / 15540
Given a sorted array of integers, find the starting and ending position of a given target value.
Your algorithm's runtime complexity must be in the order of O(log
n).
If the target is not found in the array, return [-1, -1].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4]
public class Solution { public int[] searchRange(int[] A, int target) { int[] res = {-1,-1}; if(A==null||A.length==0) return res; int start = 0,end = A.length-1; if(A[0]==target) res[0]=0; else{ while(start<=end){ int mid = start+(end-start)/2; if(mid-1>=0 && A[mid-1]<target && A[mid]==target){ res[0]=mid; break; }else if(A[mid]>=target) end = mid-1; else start = mid+1; } } start = 0; end = A.length-1; if(A[A.length-1]==target) res[1]=A.length-1; else{ while(start<=end){ int mid = start+(end-start)/2; if(mid+1<A.length && A[mid+1]>target && A[mid]==target){ res[1]=mid; break; }else if(A[mid]>target) end = mid-1; else start = mid+1; } } return res; } }